Question

For the function y=4-2x^2+1/6x^4 find the following:
Domain
x and y intercepts
Vertical asymptotes
Horizontal asymptotes
Symmetry
F'(x)
Critical numbers
Increasing f(x)
Decreasing f(x)
Extrema
F"(x)
Possible points of inflection
Concave up
Concave down
Points of inflection

I got some:
domain: (-infinity, infinity)
y-intercept: y=4
Vertical asymptotes: none
Horizontal asymptotes: none
Symmetry: y-axis
F'(x): 2/3x^3-4x
Critical numbers: x=0, x=sqrt6, x=-sqrt6
Extrema: relative maximum f(0)=4, relative minimums f(-sqrt2)=-2 and f(sqrt2)=-2
F"(x): 2x^2-4
Possible points of inflection: (sqrt2, 2/3) (-sqrt2, 2/3)
Concave up: (-infinity, -sqrt2) and (sqrt2, infinity)
Concave down: (-sqrt2, sqrt2)

To make it clearer, I can't figure out
x-intercept
Increasing f(x)
Decreasing f(x)
Points of inflection

I appreciate your time very much. Thank you.

Answer 1

taking your typing at face value the way you typed it
your domain is correct
as a matter of fact:
"I got some:
domain: (-infinity, infinity)
y-intercept: y=4
Vertical asymptotes: none
Horizontal asymptotes: none
Symmetry: y-axis
F'(x): 2/3x^3-4x
Critical numbers: x=0, x=sqrt6, x=-sqrt6
Extrema: relative maximum f(0)=4, relative minimums f(-sqrt2)=-2 and f(sqrt2)=-2
F"(x): 2x^2-4
Possible points of inflection: (sqrt2, 2/3) (-sqrt2, 2/3)
Concave up: (-infinity, -sqrt2) and (sqrt2, infinity)
Concave down: (-sqrt2, sqrt2) "

all looks good up to here

for x-intercept , we let y = 0
4-2x^2+1/6x^4 = 0
times 6
24 - 12x^2 + x^4 = 0
let's complete the square:
x^4 - 12x^2 + 36 = -24+36
(x^2 - 6)^2 = 12
x^2 - 6 = ± √12
x^2 = 6 ±√12
x = ±√(6 ±√12) , so we have 4 x-intercepts

pts of inflection, f '' (x) = 0
2x^2 - 4 = 0
x^2 = 2
x = ±√2
if x = √2 , y = 4-4+4/6 = 2/3
if x = -√2, y = 2/3
(you had that correct)

for increasing function:
f ' (x) > 0
(2/3)x^3 - 4x > 0
2x^3 - 12x > 0
x^3 - 6x > 0
x(x^2 - 6) > 0
x(x + √6)(x - √6) > 0
which is true for -√6 < x < 0 OR x > √6

it then follows that the function would be decreasing for
x < -√6 or 0 < x < √6

here is a picture of your graph:
http://www.wolframalpha.com/input/?i=plot+y%3D4-2x%5E2%2B1%2F6x%5E4